Problem: An engineer invested $\$10,\!000$ in a six-month savings certificate that paid a simple annual interest rate of $12\%$. After six months, she invested the total value of her investment in another six-month certificate. After six more months, the investment was worth $\$11,\!130$. If the annual interest rate of the second certificate is $r\%,$ then what is $r?$
For the first six months, the (simple) interest rate is $12/2 = 6$ percent.  Therefore, the investment grows to $10000 \cdot 1.06 = 10600$.

Let the annual interest rate of the second certificate be $r$ percent.  Then the interest rate for six months is $r/2$, so the investment grows to $10600 \cdot \left( 1 + \frac{r/2}{100} \right)$.  Therefore, \[10600 \cdot \left( 1 + \frac{r/2}{100} \right) = 11130.\] Then \[1 + \frac{r/2}{100} = \frac{11130}{10600} = 1.05,\] so $r/200 = 0.05$, which means $r = \boxed{10}$.